A cylindrical rod having temperature `T_(1)` and `T_(2)` at its ends. The rate of flow of heat is `Q_(1) cal//sec`. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat `Q_(2)` will be

Options

A. `4Q_1`
B. `2Q_2`
C. `Q_1/4`
D. `Q_1/2`

Answer

B

Explanation

Heat flow rate =`(KA(T_1-T_2))/L=Q`
when linear dimensions are doubled
`A_1 prop r_1^2 " " L_1=L`
`A_2 prop 4r_1^2 " " L_1=2L_1` so `Q_2=2Q_1`